There are a few different ways of answering this question.
First, I should point out that there's a mathematical machine which takes a Euclidean path integral and spits out a Hilbert space with an algebra of field operators acting on it. So if you can write down the path integral for $\phi^4$ theory rigorously, you can turn a crank and get the Hilbert space. If this theory is interacting, then the Hilbert space and algebra of operators you get won't be isomorphic to the free field Hilbert space & field algebra. (The Hilbert spaces will be isomorphic -- although not canonically -- but the field operators will be defined on different domains, and the isomorphisms won't match these domains.)
The problem with this answer is that no one has constructed the path integral for 4d $\phi^4$ theory rigorously.
In fact, there are good reasons to suspect that the continuum $\phi^4$ theory in 4d doesn't exist. The theory appears to have a Landau pole; this means that if you start with some cutoff theory and try to remove the cutoff while holding the long distance correlation functions fixed, you discover that the $\phi^4$ coupling blows up to infinity at a finite cutoff scale. Which means that there's no consistent way of turning on a $\phi^4$ interaction in pure 4d scalar field theory.
An alternate way of saying this: It looks like the $\phi^4$ interaction is marginally irrelevant; $\phi^4$ theory looks just like free scalar field theory at long distances. The interactions go away. What this means in practice for the Hilbert spaces is that they're the same.
This doesn't mean that pure $\phi^4$ theory is useless, by the way. It can still make sense as an effective field theory approximation to some short distance theory. All it means is that the short distance theory can't be pure
$\phi^4$ theory; there must be some new degrees of freedom in the UV completion.